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# Integration, Average Behavior: Extending the Definition of Integrability

Module by: Lawrence Baggett. E-mail the author

Summary: We now wish to extend the definition of the integral to a wider class of functions, namely to some that are unbounded and Others whose domains are not closed and bounded intervals. This extended definition is somewhat ad hoc, and these integrals are sometimes called “improper integrals.”

We now wish to extend the definition of the integral to a wider class of functions, namely to some that are unbounded and Others whose domains are not closed and bounded intervals. This extended definition is somewhat ad hoc, and these integrals are sometimes called “improper integrals.”

Definition 1:

Let ff be a real or complex-valued function on the open interval (a,b)(a,b) where aa is possibly -- and bb is possibly +.+. We say that ff is improperly-integrable on (a,b)(a,b) if it is integrable on each closed and bounded subinterval [a',b'](a,b),[a',b'](a,b), and for each point c(a,b)c(a,b) we have that the two limits limb'b-0cb'flimb'b-0cb'f and lima'a+0a'cflima'a+0a'cf exist.

More generally, We say that a real or complex-valued function f,f, not necessarily defined on all of the open interval (a,b),(a,b), is improperly-integrable on (a,b)(a,b) if there exists a partition {xi}{xi} of [a,b][a,b] such that ff is defined and improperly-integrable on each open interval (xi-1,xi).(xi-1,xi).

We denote the set of all functions ff that are improperly-integrable on an open interval (a,b)(a,b) by Ii((a,b)).Ii((a,b)).

Analogous definitions are made for a function's being integrable on half-open intervals [a,b)[a,b) and (a,b].(a,b].

Note that, in order for ff to be improperly-integrable on an open interval, we only require ff to be defined at almost all the points of the interval, i.e., at every point except the endpoints of some partition.

## Exercise 1

1. Let ff be defined and improperly-integrable on the open interval (a,b).(a,b). Show that lima'a+0a'cf+limb'b-0cb'flima'a+0a'cf+limb'b-0cb'f is the same for all c(a,b).c(a,b).
2. Define a function ff on (0,1)(0,1) by f(x)=(1-x)-1/2.f(x)=(1-x)-1/2. Show that ff is improperly-integrable on (0,1)(0,1) and that ff is not bounded. (Compare this with part (1) of (Reference).)
3. Define a function gg on (0,1)(0,1) by g(x)=(1-x)-1.g(x)=(1-x)-1. Show that gg is not improperly-integrable on (0,1),(0,1), and, using part (b), conclude that the product of improperly-integrable functions on (0,1)(0,1) need not itself be improperly-integrable. (Compare this with part (3) of (Reference).)
4. Define hh to be the function on (0,)(0,) given by h(x)=1h(x)=1 for all x.x. Show that hh is not improperly-integrable on (0,).(0,). (Compare this with parts (4) and (5) of (Reference).)

Part (a) of the preceding exercise is just the consistency condition we need in order to make a definition of the integral of an improperly-integrable function over an open interval.

Definition 2:

Let ff be defined and improperly-integrable on an open interval (a,b).(a,b). We define the integral of ff over the interval (a,b),(a,b), and denote it by abf,abf, by

a b f = lim a ' a + 0 a ' c f + lim b ' b - 0 c b ' f . a b f = lim a ' a + 0 a ' c f + lim b ' b - 0 c b ' f .
(1)

In general, if ff is improperly-integrable over an open interval, i.e., ff is defined and improperly-integrable over each subinterval of (a,b)(a,b) determined by a partition {xi},{xi}, then we define the integral of ff over the interval (a,b)(a,b) by

a b f = i = 1 n x i - 1 x i f . a b f = i = 1 n x i - 1 x i f .
(2)

## Theorem 1

Let (a,b)(a,b) be a fixed open interval (with aa possibly equal to -- and bb possibly equal to +),+), and let Ii((a,b))Ii((a,b)) denote the set of improperly-integrable functions on (a,b).(a,b). Then:

1.  Ii((a,b))Ii((a,b)) is a vector space of functions.
2.  (Linearity) ab(αf+βg)=αabf+βabgab(αf+βg)=αabf+βabg for all f,gIi((a,b))f,gIi((a,b))and α,βC.α,βC.
3.  (Positivity) If f(x)0f(x)0 for all x(a,b),x(a,b), then abf0.abf0.
4.  (Order-preserving) If f,gIi((a,b))f,gIi((a,b)) and f(x)g(x)f(x)g(x) for all x(a,b),x(a,b), then abfabg.abfabg.

## Exercise 2

1. Use (Reference), (Reference), (Reference), and properties of limits to prove the preceding theorem.
2. Let ff be defined and improperly-integrable on (a,b).(a,b). Show that, given an ϵ>0,ϵ>0, there exists a δ>0δ>0 such that for any a<a'<a+δa<a'<a+δ and any b-δ<b'<bb-δ<b'<b we have |aa'f|+|b'bf|<ϵ.|aa'f|+|b'bf|<ϵ.
3. Let ff be improperly-integrable on an open interval (a,b).(a,b). Show that, given an ϵ>0,ϵ>0, there exists a δ>0δ>0 such that if (c,d)(c,d) is any open subinterval of (a,b)(a,b) for which d-c<δ,d-c<δ, then |cdf|<ϵ.|cdf|<ϵ. HINT: Let {xi}{xi} be a partition of [a,b][a,b] such that ff is defined and improperly-integrable on each subinterval (xi-1,xi).(xi-1,xi). For each i,i, choose a δiδi using part (b). Now ff is bounded by MM on all the intervals [xi-1+δi,xi-δi],[xi-1+δi,xi-δi], so δ=ϵ/Mδ=ϵ/M should work there.
4. Suppose ff is a continuous function on a closed bounded interval [a,b][a,b] and is continuously differentiable on the open interval (a,b).(a,b). Prove that f'f' is improperly-integrable on (a,b),(a,b), and evaluate abf'.abf'. HINT: Fix a point c(a,b),c(a,b), and use the Fundamental Theorem of Calculus to show that the two limits exist.
5. (Integration by substitution again.) Let g:[c,d][a,b]g:[c,d][a,b] be continuous on [c,d][c,d] and satisfy g(c)=ag(c)=a and g(d)=b.g(d)=b. Suppose there exists a partition {x0<x1<...<xn}{x0<x1<...<xn} of the interval [c,d][c,d] such that gg is continuously differentiable on each subinterval (xi-1,xi).(xi-1,xi). Prove that g'g' is improperly-integrable on the open interval (c,d).(c,d). Show also that if ff is continuous on [a,b],[a,b], we have that
abf(t)dt=cdf(g(s))g'(s)ds.abf(t)dt=cdf(g(s))g'(s)ds.
(3)
HINT: Integrate over the subintervals (xi-1,xi),(xi-1,xi), and use part (d).

REMARK Note that there are parts of (Reference) and (Reference) that are not asserted in Theorem 1. The point is that these other properties do not hold for improperly-integrable functions on open intervals. See the following exercise.

## Exercise 3

1. Define ff to be the function on [1,)[1,) given by f(x)=(-1)n-1/nf(x)=(-1)n-1/n if n-1x<n.n-1x<n. Show that ff is improperly-integrable on (1,),(1,), but that |f||f| is not improperly-integrable on (1,).(1,). (Compare this with part (4) of (Reference).) HINT: Verify that 1Nf1Nf is a partial sum of a convergent infinite series, and then verify that 1N|f|1N|f| is a partial sum of a divergent infinite series.
2. Define the function ff on (1,)(1,) by f(x)=1/x.f(x)=1/x. For each positive integer n,n, define the function fnfn on (1,)(1,) by fn(x)=1/xfn(x)=1/x if 1<x<n1<x<n and fn(x)=0fn(x)=0 otherwise. Show that each fnfn is improperly-integrable on (1,),(1,), that ff is the uniform limit of the sequence {fn},{fn}, but that ff is not improperly-integrable on (1,).(1,). (Compare this with part (5) of Theorem 5.6.)
3. Suppose ff is a nonnegative real-valued function on the half-open interval (a,)(a,) that is integrable on every closed bounded subinterval [a,b'].[a,b']. For each positive integer na,na, define yn=anf(x)dx.yn=anf(x)dx. Prove that ff is improperly-integrable on [a,)[a,) if and only if the sequence {yn}{yn} is convergent. In that case, show that af=limyn.af=limyn.

We are now able to prove an important result relating integrals over infinite intervals and convergence of infinite series.

## Theorem 2

Let ff be a positive function on [1,),[1,), assume that ff is integrable on every closed bounded interval [1,b],[1,b], and suppose that ff is nonincreasing; i.e., if x<yx<y then f(x)f(y).f(x)f(y). For each positive integer i,i, set ai=f(i),ai=f(i), and let SNSN denote the NNth partial sum of the infinite series ai:ai:SN=i=1Nai.SN=i=1Nai. Then:

1. For each N,N, we have
SN-a11Nf(x)dxSN-1.SN-a11Nf(x)dxSN-1.
(4)
2. For each N,N, we have that
SN-1-1Nf(x)dxa1-aNa1;SN-1-1Nf(x)dxa1-aNa1;
(5)
i.e., the sequence {SN-1-1Nf}{SN-1-1Nf} is bounded above.
3. The sequence {SN-1-1Nf}{SN-1-1Nf} is nondecreasing.
4.  (Integral Test) The infinite series aiai converges if and only if the function ff is improperly-integrable on (1,).(1,).

### Proof

For each positive integer N,N, define a step function kNkN on the interval [1,N][1,N] as follows. Let P={x0<x1<...<xN-1}P={x0<x1<...<xN-1} be the partition of [1,N][1,N] given by the points {1<2<3<...<N},{1<2<3<...<N}, i.e., xi=i+1.xi=i+1. Define kN(x)kN(x) to be the constant ci=f(i+1)ci=f(i+1) on the interval [xi-1,xi)=[i,i+1).[xi-1,xi)=[i,i+1). Complete the definition of kNkN by setting kN(N)=f(N).kN(N)=f(N). Then, because ff is nonincreasing, we have that kN(x)f(x)kN(x)f(x) for all x[1,N].x[1,N]. Also,

1 N k N = i = 1 N - 1 c i ( x i - x i - 1 ) = i = 1 N - 1 f ( i + 1 ) = i = 2 N f ( i ) = i = 2 N a i = S N - a 1 , 1 N k N = i = 1 N - 1 c i ( x i - x i - 1 ) = i = 1 N - 1 f ( i + 1 ) = i = 2 N f ( i ) = i = 2 N a i = S N - a 1 ,
(6)

which then implies that

S N - a 1 = 1 N k N ( x ) d x 1 N f ( x ) d x . S N - a 1 = 1 N k N ( x ) d x 1 N f ( x ) d x .
(7)

This proves half of part (1).

For each positive integer N>1N>1 define another step function lN,lN, using the same partition PP as above, by setting lN(x)=f(i)lN(x)=f(i) if ix<i+1ix<i+1 for 1i<N,1i<N, and complete the definition of lNlN by setting lN(N)=f(N).lN(N)=f(N). Again, because ff is nonincreasing, we have that f(x)lN(x)f(x)lN(x) for all x[1,N].x[1,N]. Also

1 N l N = i = 1 N - 1 f ( i ) = i = 1 N - 1 a i = S N - 1 , 1 N l N = i = 1 N - 1 f ( i ) = i = 1 N - 1 a i = S N - 1 ,
(8)

which then implies that

1 N f ( x ) d x 1 N l N ( x ) d x = S N - 1 , 1 N f ( x ) d x 1 N l N ( x ) d x = S N - 1 ,
(9)

and this proves the other half of part (1).

It follows from part (1) that

S N - 1 - 1 N f ( x ) d x S N - 1 - S N + a 1 = a 1 - a N , S N - 1 - 1 N f ( x ) d x S N - 1 - S N + a 1 = a 1 - a N ,
(10)

and this proves part (2).

We see that the sequence {SN-1-1Nf}{SN-1-1Nf} is nondecreasing by observing that

S N - 1 N + 1 f - S N - 1 + 1 N f = a N - N N + 1 f = f ( N ) - N N + 1 f 0 , S N - 1 N + 1 f - S N - 1 + 1 N f = a N - N N + 1 f = f ( N ) - N N + 1 f 0 ,
(11)

because ff is nonincreasing.

Finally, to prove part (4), note that both of the sequences {SN}{SN} and {1Nf}{1Nf} are nondecreasing. If ff is improperly-integrable on [1,),[1,), then limN1NflimN1Nf exists, and SNa1+1f(x)dxSNa1+1f(x)dx for all N,N, which implies that aiai converges by (Reference). Conversely, if aiai converges, then limSNlimSN exists. Since 1Nf(x)dxSN-1,1Nf(x)dxSN-1, it then follows, again from (Reference), that limN1Nf(x)dxlimN1Nf(x)dx exists. So, by the preceding exercise, ff is improperly-integrable on [1,).[1,).

We may now resolve a question first raised in (Reference). That is, for 1<s<2,1<s<2, is the infinite series 1/ns1/ns convergent or divergent? We saw in that exercise that this series is convergent if ss is a rational number.

## Exercise 4

1. Let ss be a real number. Use the Integral Test to prove that the infinite series 1/ns1/ns is convergent if and only if s>1.s>1.
2. Let ss be a complex number s=a+bi.s=a+bi. Prove that the infinite series 1/ns1/ns is absolutely convergent if and only if a>1.a>1.

## Exercise 5

Let ff be the function on [1,)[1,) defined by f(x)=1/x.f(x)=1/x.

1. Use Theorem 2 to prove that the sequence {i=1N1i-lnN}{i=1N1i-lnN} converges to a positive number γ1.γ1. (This number γγ is called Euler's constant.) HINT: Show that this sequence is bounded above and nondecreasing.
2. Prove that
i=1(-1)i+1i=ln2.i=1(-1)i+1i=ln2.
(12)
HINT: Write S2NS2N for the 2N2Nth partial sum of the series. Use the fact that
S2N=i=12N1i-2i=1N12i.S2N=i=12N1i-2i=1N12i.
(13)
Now add and subtract ln(2N)ln(2N) and use part (a).

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